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Unformatted text preview: Miller, Kierste Exam 2 Due: Mar 27 2007, 11:00 pm Inst: Gary Berg 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The region R in the first quadrant bounded by the xaxis and the graph of y = ln(1 + 4 x 2 x 2 ) is shown in R 1 2 Estimate the area of R using Simpsons Rule with 2 equal subintervals. 1. area ( R ) 2 ln 5 2 2. area ( R ) 4 3 ln 3 correct 3. area ( R ) 4 3 ln 2 4. area ( R ) 2 ln 3 5. area ( R ) 4 3 ln 5 2 6. area ( R ) 2 ln 2 Explanation: The area of R is given by the definite inte gral I = Z 2 ln(1 + 4 x 2 x 2 ) dx. Now by Simpsons Rule with 2 equal subin tervals, I h 3 n f (0) + 4 f (1) + f (2) o , h = 1 , where f (0) = 0 , f (1) = ln 3 , f (2) = 0 . Consequently, area R 4 3 ln 3 . keywords: Stewart5e, numerical integration, Simpsons Rule, log function, estimate area 002 (part 1 of 1) 10 points Evaluate the integral I = Z / 4 sec 2 x { 3 2 sin x } dx. 1. I = 5 + 2 2. I = 5 + 2 2 3. I = 1 2 4. I = 5 2 2 correct 5. I = 1 2 2 6. I = 1 + 2 Explanation: Since sec 2 x { 3 2 sin x } = 3 sec 2 x 2 sec x sin x cos x , we see that I = Z / 4 { 3 sec 2 x 2 sec tan x } dx. But d dx tan x = sec 2 x, Miller, Kierste Exam 2 Due: Mar 27 2007, 11:00 pm Inst: Gary Berg 2 while d dx sec x = sec x tan x. Consequently, I = h 3 tan x 2 sec x i / 4 = 5 2 2 . keywords: definite integral, tan integral, sec integral 003 (part 1 of 1) 10 points Evaluate the integral I = Z 1 1 x 2 + 1 dx. 1. I = 2 ( 2 1 ) 2. I = 2 1 3. I = 2 ln(1 + 2 ) 4. I = 2 (1 + 2 ) 5. I = ln(1 + 2 ) correct 6. I = ln( 2 1 ) Explanation: Set x = tan u, then dx = sec 2 udu, x 2 + 1 = sec 2 u, while x = 0 = u = 0 , x = 1 = u = 4 . In this case I = Z / 4 sec 2 u sec u du = Z / 4 sec udu. On the other hand, Z sec udu = ln  sec u + tan u  + C . Thus I = ln  sec u + tan u  / 4 . Consequently, I = ln(1 + 2 ) . keywords: 004 (part 1 of 1) 10 points Evaluate the integral I = Z e x sin xdx. 1. I = 1 2 (1 e ) 2. I = e + 1 3. I = e  1 4. I = 1 e 5. I = 1 2 ( e + 1) 6. I = 1 2 ( e + 1) correct 7. I = e + 1 8. I = 1 2 ( e + 1) Explanation: After integration by parts, I = h e x sin x i  Z e x cos xdx = 0 Z e x cos xdx....
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This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Cepparo

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